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4 years ago

What are the steps to naming a covalent compound?

1 answer:

8 0

First you identify the elements present second look at the subscript of each element to determine which prefix to use and third apply the above naming scheme

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Which of the following manifests the harmonious bringing forth of technology?

irinina [24]

Answer:

iPhones

Explanation:

Hope this helps:)

3 0

3 years ago

2HgO—>2Hg+O2. I start with 46.8 g HgO. What is my theoretical yield of O2?

jok3333 [9.3K]

Answer:

Theoretical yield = 3.52 g

Percent yield =65.34%

Explanation:

Given data:

Mass of HgO = 46.8 g

Theoretical yield of O₂ = ?

Percent yield of O₂ = ?

Actual yield of O₂ = 2.30 g

Solution:

Chemical equation:

2HgO → 2Hg + O₂

Number of moles of HgO = mass/ molar mass

Number of moles of HgO = 46.8 g / 216.6 g/mol

Number of moles of HgO = 0.22 mol

Now we will compare the moles of HgO with oxygen.

HgO : O₂

2 : 1

0.22 : 1/2×0.22 = 0.11 mol

Theoretical yield:

Mass of oxygen = number of moles × molar mass

Mass of oxygen = 0.11 mol × 32 g/mol

Mass of oxygen = 3.52 g

Percent yield :

Percent yield = actual yield / theoretical yield × 100

Percent yield = 2.30 g/ 3.52 g × 100

Percent yield =65.34%

5 0

3 years ago

Consider a hexagonal close-packed unit cell as shown here.

Alenkinab [10]

Sorry, but I can't help unless you post a diagram or picture. If you can't, what about a written description?

7 0

4 years ago

The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it

DedPeter [7]

Answer:

It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.

Explanation:

Radioactive decays/reactions always follow a first order reaction dynamic.

Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A

The general expression for rate of reaction for a first order reaction is

(dA/dt) = -kA (Minus sign because it's a rate of reduction)

k = rate constant

(dA/dt) = -kA

(dA/A) = -kdt

∫ (dA/A) = -k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.

We get

In (A/A₀) = -kt

(A/A₀) = e⁻ᵏᵗ

A(t) = A₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 5730 years

k = (In 2)/5730 = 0.000120968 = 0.000121 /year.

So, the amount of C-14 atoms left at any time is given as

A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ

How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?

When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.

Hence,

A(t) = 75%

A₀ = 100%

100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ

e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75

In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768

-0.000121t = -0.28768

t = (0.28768/0.000121) = 2,377.54 = 2378 years

Hope this Helps!!!

3 0

4 years ago

The combustion of caffeine with the molecular masses is given below. If you have 0.150 grams of caffeine, how much NO2 in grams

lord [1]

Answer:

1. 0.14 g of NO2.

2. 0.27 g of CO2.

Explanation:

The balanced equation for the reaction is given below:

2C8H10N4O2 + 27O2 —> 16CO2 + 10H2O + 8NO2

Next, we shall determine the mass of caffeine, C8H10N4O2 that reacted and the masses of nitrogen (iv) oxide, NO2 and carbon (iv) oxide, CO2 produced from the balanced equation. This can be obtained as follow:

Molar mass of of C8H10N4O2 = 194.19 g/mol

Mass of C8H10N4O2 from the balanced equation = 2 × 194.19 = 388.38 g

Molar mass of CO2 = 44.01 g/mol

Mass of CO2 from the balanced equation = 16 × 44.01 = 704.16 g

Molar mass of NO2 = 46.01 g/mol

Mass of NO2 from the balanced equation = 8 × 46.01 = 368.08 g

Summary:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 704.16 g of CO2 and 368.08 g of NO2.

1. Determination of the mass of NO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 368.08 g of NO2.

Therefore, 0.15 g of caffeine, C8H10N4O2, will react to produce = (0.15 × 368.08) / 388.38 = 0.14 g of NO2.

Therefore, 0.14 g of NO2 was obtained from the reaction.

2. Determination of the mass of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 704.16 g of CO2.

Therefore, 0.15 g of caffeine, C8H10N4O2, will react to produce = (0.15 × 704.16) / 388.38 = 0.27 g of CO2.

Therefore, 0.27 g of CO2 was obtained from the reaction.

8 0

3 years ago